Integrand size = 38, antiderivative size = 117 \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {i B \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}-\frac {i B \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d} \]
I*B*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/(I*a-b )^(1/2)-I*B*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2)) /d/(I*a+b)^(1/2)
Time = 0.06 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt [4]{-1} B \left (-\frac {\arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )}{d} \]
((-1)^(1/4)*B*(-(ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqr t[a + b*Tan[c + d*x]]]/Sqrt[-a + I*b]) + ArcTan[((-1)^(1/4)*Sqrt[a + I*b]* Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/Sqrt[a + I*b]))/d
Time = 0.33 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {2011, 3042, 4058, 613, 104, 218, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 2011 |
\(\displaystyle B \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4058 |
\(\displaystyle \frac {B \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 613 |
\(\displaystyle \frac {B \left (\frac {1}{2} \int \frac {1}{\sqrt {\tan (c+d x)} (\tan (c+d x)+i) \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)-\frac {1}{2} \int \frac {1}{(i-\tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {B \left (\int \frac {1}{\frac {(a-i b) \tan (c+d x)}{a+b \tan (c+d x)}+i}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}-\int \frac {1}{i-\frac {(a+i b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {B \left (\int \frac {1}{\frac {(a-i b) \tan (c+d x)}{a+b \tan (c+d x)}+i}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}+\frac {i \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}\right )}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {B \left (\frac {i \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}-\frac {i \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b+i a}}\right )}{d}\) |
(B*((I*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] )/Sqrt[I*a - b] - (I*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b *Tan[c + d*x]]])/Sqrt[I*a + b]))/d
3.5.70.3.1 Defintions of rubi rules used
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[Sqrt[(e_.)*(x_)]/(Sqrt[(c_) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^2)), x_Sym bol] :> Simp[e/(2*b) Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] + x)), x ], x] - Simp[e/(2*b) Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] - x)), x ], x] /; FreeQ[{a, b, c, d, e}, x]
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S imp[ff/f Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a *d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 0.57 (sec) , antiderivative size = 940499, normalized size of antiderivative = 8038.45
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 4509 vs. \(2 (89) = 178\).
Time = 0.75 (sec) , antiderivative size = 4509, normalized size of antiderivative = 38.54 \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
-1/8*sqrt(((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 - B^2*b)/((a^2 + b^2)*d^2))*log((2*(2*B^3*a^4*b + 4*B^3*a^2*b^3 + (B^3*a^5 + 3*B^3*a^3*b^2 + 4*B^3*a*b^4)*tan(d*x + c) + (2*(B*a^5*b + 3*B*a^3*b^3 + 2 *B*a*b^5)*d^2*tan(d*x + c) - (B*a^6 + 4*B*a^4*b^2 + 7*B*a^2*b^4 + 4*B*b^6) *d^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + ((B^2*a^6 + 7*B^2*a^4*b^2 + 12*B^2*a^2*b^4)*d*tan( d*x + c)^2 + 2*(B^2*a^5*b + B^2*a^3*b^3 - 4*B^2*a*b^5)*d*tan(d*x + c) - (B ^2*a^6 + 3*B^2*a^4*b^2 + 4*B^2*a^2*b^4)*d - 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^ 7)*d^3*tan(d*x + c)^2 + (a^7 + 6*a^5*b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3*tan(d *x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2*b^5)*d^3)*sqrt(-B^4*a^2/((a^4 + 2*a^2 *b^2 + b^4)*d^4)))*sqrt(((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4 )*d^4))*d^2 - B^2*b)/((a^2 + b^2)*d^2)))/(tan(d*x + c)^2 + 1)) - 1/8*sqrt( ((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 - B^2*b)/((a ^2 + b^2)*d^2))*log(-(2*(2*B^3*a^4*b + 4*B^3*a^2*b^3 + (B^3*a^5 + 3*B^3*a^ 3*b^2 + 4*B^3*a*b^4)*tan(d*x + c) + (2*(B*a^5*b + 3*B*a^3*b^3 + 2*B*a*b^5) *d^2*tan(d*x + c) - (B*a^6 + 4*B*a^4*b^2 + 7*B*a^2*b^4 + 4*B*b^6)*d^2)*sqr t(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(t an(d*x + c)) + ((B^2*a^6 + 7*B^2*a^4*b^2 + 12*B^2*a^2*b^4)*d*tan(d*x + c)^ 2 + 2*(B^2*a^5*b + B^2*a^3*b^3 - 4*B^2*a*b^5)*d*tan(d*x + c) - (B^2*a^6 + 3*B^2*a^4*b^2 + 4*B^2*a^2*b^4)*d - 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^3...
\[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=B \int \frac {\sqrt {\tan {\left (c + d x \right )}}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B b \tan \left (d x + c\right ) + B a\right )} \sqrt {\tan \left (d x + c\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (B\,a+B\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]